\(\displaystyle \blacktriangleright \;\) 
Assume by absurd that it does. Then the remark above shows that \(\displaystyle \langle \mathrm {Spin}_{7}, \mathrm {SL}_{3} \rangle = \mathrm {E}_6\) . One immediately checks that the action of \(\displaystyle \langle \mathrm {Spin}_{7}, \mathrm {SL}_{3} \rangle \)  we defined on \(\displaystyle \mathrm {J}_{3}(\mathbb {O})\)  preserves the rank, when we see elements of \(\displaystyle \mathrm {J}_{3}(\mathbb {O})\)  as \(\displaystyle 24*24\)  matrices. Since any element of \(\displaystyle \mathrm {J}_{3}(\mathbb {O})\)  can be diagonalized by the action of \(\displaystyle \mathrm {E}_6\)  (see the chapter 14 of [1]}), this would prove that any element of \(\displaystyle \mathrm {J}_{3}(\mathbb {O})\)  has rank \(\displaystyle 0,8,16\)  or \(\displaystyle 24\)  as a \(\displaystyle 24*24\)  matrix. But we know that a generic point on \(\displaystyle \mathfrak {S} = 0\)  has rank \(\displaystyle 22\)  (see Theorem REF ), a contradiction.
